# Modifying a renormalisable theory$.$

Apparently, if we take a certain renormalisable theory, then any modification consistent with the symmetries must render the theory non-renormalisable. Is this claim true? Has it been discussed rigorously in the literature?

• would the downvoter care to comment? Jul 18 '17 at 15:36
• 1. By renormalisable I mean: there exists a counter-term for any divergence that appears in perturbation theory. 2. By "a modification" I mean: adding a term to the Lagrangian (without introducing new fields). Jul 18 '17 at 15:38
• "Non-renormalizable theories are just as renormalizable as renormalizable theories, as long as we include all possible terms in the Lagrangian." S. Weinberg QFT book Vol. 1 section 12.3
– jak
Jul 19 '17 at 8:40

"Any modification" is pretty vague. The point is the following. Given a number of fields $\phi_I$ (that can be bosons, fermions, gauge fields, transforming under some internal symmetries etc.) you can write down a finite number of local, gauge-invariant operators of dimension $\leq 4$. Call these $\mathcal{O}_\alpha$. Then the most general renormalizable action reads $$L = \text{kinetic terms} + \sum_{\alpha} g_\alpha \mathcal{O}_\alpha$$ where the couplings $g_\alpha$ have mass dimension $4 - \text{dimension of } \mathcal{O}_\alpha$.
You can of course write down many more operators, of dimension $5,6,\ldots$. In principle we can add such operators, with couplings $g'_\beta$, to the action as well. So a point in theory space is parametrized by an infinite vector $(g_\alpha, g'_\beta)$. The point is that the submanifold with $g'_\beta = 0$ corresponds to the set of renormalizable trajectories. This is what that David Bar Moshe was aiming at. In the case of Yang-Mills, there is only one coupling $g_\alpha = g_{YM}$, so any other operator you add to the action will destroy renormalizability.
• so given the finite set $\{g_\alpha\}=\{g_1,\cdots,g_n\}$, if I artificially remove the last coupling, $g_n\equiv 0$, does the theory become non-renormalisable? [this doesn't even seem to be true for $g_1\phi^3+g_2\phi^4$ theory, where I can set $g_2=0$ and the theory remains renormalisable...] Jul 18 '17 at 15:35
• No, it doesn't become non-renormalizable. But you may have to add the $g_n$ coupling as a counterterm. For instance, in $\phi^4$ theory the low-energy physics is governed by a (renormalized) mass $m$ and a coupling $g$. The bare values $m_0, g_0$ are dictated by these. Or you can think of this differently: if $g_0 > 0$ you have to add a non-zero bare mass, otherwise you cannot take the cutoff to infinity.